Question

Q
Q.20
To a given container having a pore of definite
size, gas A (mol. wt = 81) is filled till the final
pressure become 10 atm. It was seen in 50
minutes 10 gm of A was effused out. Now the
container was completely evacuated and filled
with gas B (mol. wt = 100) till the final
pressure become 20 atm. In 75 minutes how
many gm of B will be effused out ?
(B) 100/6 gm
(A) 100/3 gm

(© 200/3 gm
(D) 250/3 gm​

Answer 1

answer : option (c) 200/3 gm

explanation :

case 1: $P_A=10atm$, $M_A=81$ , $t_A=50min$ and $m_A=10gm$

so, number of mole of gas A was effused, $n_A$= (10/81)

so, rate of effusion of gas A = $\frac{V_B}{t_A}=kP_A\frac{1}{\sqrt{M_A}}$

or, $\frac{V_A}{50}=k(10)\frac{1}{\sqrt{81}}=\frac{10k}{9}$.....(1)

where $V_A$ is volume of gas effused in given time and k is proportionality constant.

case 2 : $P_2=20atm$, $t_B=75min$,$M_B=100$

so, rate of effusion of gas B = $\frac{V_B}{t_B}=kP_B\frac{1}{\sqrt{M_B}}$

or, $\frac{V_B}{75}=k(20)\frac{1}{\sqrt{100}}=2k$.....(2)

from equations (1) and (2),

$\frac{75V_A}{50V_B}=\frac{10}{18}$

or, $\frac{V_A}{V_B}=\frac{10}{27}$

now applying $\frac{P_AV_A}{n_A}=\frac{P_BV_B}{n_B}$

or, 10 × 10/(10/81) = 20 × 27/(x/100)

or, x = 200/3

hence, 200/3 gm of gas B will be effused out .